Here's the problem: indefinite integral of (-3x^2)/sqrt((t^3)-16)
If you are unsure of your result, use differentiation to check your result:
$\displaystyle \frac{d}{dx}\left(\frac{\sqrt{t^3+16}}{2}+C \right)=\frac{3t^2}{4\sqrt{t^3+16}}\ne\frac{-3t^2}{\sqrt{t^3+16}}$
We can see it is close (essentially the wrong constant factor of the radical), but not quite right. I would let:
$\displaystyle u=t^3+16\,\therefore\,du=3t^2\,dt$ and so we have:
$\displaystyle -\int u^{-\frac{1}{2}}\,du=-2u^{\frac{1}{2}}+C=-2\sqrt{t^3+16}+C$
Now, checking by differentiation, we find:
$\displaystyle \frac{d}{dx}\left(-2\sqrt{t^3+16}+C \right)=\frac{-3t^2}{2\sqrt{t^3+16}}$
The derivative of the anti-derivative is the original integrand, so our result is correct.